MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $E$ and $F$ be $n$-dimensional representations of $S_n$ of the form $V\oplus\mathbb{C}$, where $V$ is the standard representation. I just wonder if there is a formula to decompose $S^k(E\otimes F)\otimes\Lambda^m(E\otimes F)$ into irreducible modules of $S_n\times S_n$. Many thanks.

share|cite|improve this question
up vote 4 down vote accepted

This is very far from a complete answer, but it should give enough results to enable the irreducible constituents to be computed for some small values of $k$, $m$ and $n$. I would be amazed if there was an easy formula for the decomposition. To clarify the notation, I'll write $\boxtimes$ for the outer tensor product, so if $U$ and $V$ are representation of $S_n$ then $U \boxtimes V$ is a representation of $S_n \times S_n$.

The question asks for the inner tensor product of $S^k( E\boxtimes F)$ and $\Lambda^m(E \boxtimes F)$ where $E$ and $F$ are the natural $n$-dimensional permutation representations of $S_n$. The symmetric power and exterior powers can be decomposed using the plethystic identities

$$ S^k(E \boxtimes F) = \sum_\mu \Delta^\mu(E) \boxtimes \Delta^\mu (F)$$

$$ \Lambda^m (E \boxtimes F) = \sum_\mu \Delta^\mu(E) \boxtimes \Delta^{\mu'}(F)$$

where the sums all over all partitions $\mu$ of $k$ or $m$ respectively, $\Delta^\mu(E)$ is the image of $E$ under the Schur functor for the partition $\mu$, and $\mu'$ is the conjugate partition to $\mu$. One source for these identities is this article by Loehr and Remmel (see bottom of page 191). They are generalizations of the isomorphism of Schur functors used by S. Carnahan in this Mathoverflow answer.

We then need to understand $\Delta^\mu(E)$ as a representation of $S_n$. I believe this is very tricky, and only a few special cases are known. If $\mu = (k)$ then $\Delta^k(E) = S^k E$: in this case the character can be written as a sum of Young permutation characters, and the irreducible constituents determined by Young's rule. (There are also relevant results from invariant theory.) If $\mu = (1^k)$ then $\Delta^k(E) = \Lambda^k(E)$ and the character is well known to be

$$ \chi_{\Lambda^k(E)} = \chi^{(n-k+1,1^{k-1})} + \chi^{(n-k,1^k)} $$

using the standard notation.

Assuming that the characters of $S^k(E\boxtimes F)$ and $\Lambda^m(E \boxtimes F)$ have been expressed as sums of irreducible characters of $S_n \times S_n$, the identity $$ (U \boxtimes V) \otimes (U' \boxtimes V') \cong (U \otimes U') \boxtimes (V \otimes V') $$ then reduces the problem to decomposing various inner tensor products of representations of $S_n$ into irreducible representations. This is yet another very tricky problem and few general results are known. This paper by Bessenrodt and Kleshchev is a good introduction. One very special case is tensor products with the $(n-1)$-dimensional standard representation: for this the identity

$$ \chi^\lambda \chi^{(n-1,1)} = \rm{Ind}^{S_n} \bigl(\rm{Res}_{S_n-1} \chi^\lambda \bigr) - \chi^{(n-1,1)} $$

combined with the ordinary branching rule is often useful.

To give a small example, suppose we want to compute the character of $S^2(E \boxtimes F) \otimes (E \boxtimes F)$. The character of $S^2 E$ is the sum of the Young permutation characters $ \pi^{(n-2,2)} + \pi^{(n-1,1)} $ and so $$\chi_{S^2 E} = 2\chi^{(n)} + 2\chi^{(n-1,1)} + \chi^{(n-2,2)}.$$ By the result on exterior powers mentioned above, $\chi_{\Lambda^2 E} = \chi^{(n-1,1)} + \chi^{(n-2,1,1)}$. This gives enough to write $S^2(E \boxtimes F)$ as a sum of irreducible characters of $S_n \times S_n$, and then the restriction / induction trick can be used to finish the calculation.

share|cite|improve this answer
Many thanks for your help, Mark. I think I can try some examples for small k, m and n now. – JYQ Apr 14 '12 at 20:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.